All Answers 358090 Answers

x^2 + 2    cannot be factored in terms of real numbers.... if we allow for non-reals, we have

[x - (√2)i] [ [x + (√2)i ]

2^30   = 1,073,741,824

If tanA=xsinB/[1-xcosB] & tanB=ysinA/[1-ycosA] then prove that -

sinA/sinB= x/y

tanA  = sinA/cosA   and    tanB = sinB/ cosB   ......therefore, we can write :

sinA /cosA  = [xsinB]/ [1 - xcosB]   (1)

sinB/cosB  = [ysinA] / [ 1 - y cosA]  (2)

Rearranging (1)and (2), we have

[sinA] / [sinB]  = [x cosA]/ [1 - xcosB]   (3)

[sinA]/ [sinB] =  [1 - ycosA] / [y cosB]    (4)

Setting the right sides of  (3) and (4) equal gives us

[x cosA]/ [1 - xcosB]   = [1 - ycosA] / [y cosB]     separate the right side

[xcosA] / [1 - xcosB] = [ 1/[ycosB] - cosA/cosB]   cross-multiply

[xcosA]= [1 - xcosB] [1/[ycosB] - cosA/cosB]

[xcosA] = 1/[ycosB] - [x/y] - cosA/cosB + xcosA   

[x/y]   =  1/[ycosB] - [cosA/cosB]     get a common denominator on the right

[x/y]  =  [ 1 - ycosA] / [ ycosB]

And, from (4),  [1 - ycosA]/ [y cosB]  =  [sinA] / [sinB].......therefore,

x/y  =  sinA / sin B

cos(3pi/2 radians)  =   cos(270°)   =  cos(-90°)  = cos(90°)  = 0

.

See my (corrected) answer at http://web2.0calc.com/questions/if-a-b-c-pi-amp-cosa-cosb-cosc-then-prove#r2 

My reply above that part (i) is incorrect is itself incorrect!! (I forgot about the cos A = cos B*cos C bit!)

3 / 4*5

Of course it does, it means that you got 2 different answers.

20x^3

10m^4, 40m^7 and 100^m^5

The greatest common factor of 10, 40 and 100  = 10

The greatest common factor of m^4, m^7 and m^5  = m^4

So  the greatest common factor is 10m^4

One angle T is divided in two part A & B where tanA/tanB=x/y. Prove that -

sin(A-B)=(x-y)/(x+y)*sinT

sinT  = sin(A + B)

x = tan A,  y  = tan B

 (x - y)/(x + y)sin(A + B)  =  

[tanA - tanB] / [tanA + tanB] sin(A + B)  =

[(sinA/cosA) - (sinB/cosB)] / [ (sinA/cosA) + (sinB/cosB)] sin(A + B) =

[(sinAcosB - sinBcosA) / cosAcosB] / [(sinAcosB + sinBcosA)/ cosAcosB] sin(A + B) =

 ([sinAcosB - sinBcosA] /[ [sinAcosB + sinBcosA] )  *  [ sinAcosB + sinBcosA]  =

[sinAcosB - sinBcosA]  =

sin(A - B)

Very nice Melody.....!!!!.......here's the math.....

PA^2  = (AD -v)^2 + w^2

PC^2  = (CD - w)^2  + v^2

PB^2  = (CD -w)^2 + (AD - v)^2

PD^2  = v^2 + w^2

So

PA^2 + PC^2   = PB^2 + PD^2   ???

(AD -v)^2 + w^2 + (CD - w)^2  + v^2  = (CD -w)^2 + (AD - v)^2 + v^2 + w^2  ???

Rearrange the terms  on the right

(AD -v)^2 + w^2 + (CD - w)^2  + v^2  = (AD -v)^2 + w^2 + (CD - w)^2  + v^2

And it's true  !!!!

[P. S.  - I liked this one !!!  ]

A grocery store chain needs to transport of 3000m^3 of refrigerated goods and 4000m^3 of non-refrigerated goods. They plan to hire truck from a company that has two types of trucks for rent. A truck of type A has a refrigerated goods section of 20m^3 and a non-refrigerated goods section of 40m^3, while each truck of type B has both sections of the same volume of 30m^3. The cost per kilometer of a Type A is $30, and $40 for Type B. How many trucks of each type should the grocer rent to achieve the minimum total cost?

I think you must mean cost "per cubic meter" rather than cost "per kilometer."

Let x be the number of type A trucks needed and y be the number of type B trucks needed.

Here are the constraints

20x + 30y ≤ 3000

40x + 30y ≤ 4000

The total cost will be the number of  each type of truck rented* the capacity of each * the cost per cubic meter....so....we have the following objective function, z, which we wish to minimze

z = x*60*30 + y*60*40

z = 1800x + 2400y

Here's the graph of the constraints : https://www.desmos.com/calculator/mr9doouggj

The minimum cost is realized at the corner point of the intersection of both inequalities......this happens at (50, 66.667)   .......so  you need 50 trucks of type A  and 67 trucks of type B

And the minimum  cost will be $270,000

sum_(n=0)^99 ((-1)^n (pi*i)^n)/n!)~ -1.0000000000000000000000000000000000000000+0.×10^-40 i

Source: Wolfram/Alpha.

is m(metre)^-3 = mm^3 ?

No!.

1 m^-3=1×10^-9 mm^(-3) (reciprocal cubic millimeters)

No,

mm stands for millimetres and m stands for metres

There are 1000mm in a metre

So that means there are 1000^2 mm^2 in a metre^2

and there are 1000^3 mm^3 in a metre^3

so

\(1,000,000,000 mm^3=1m^3\\ or\\ 1mm^3=10^{-9}m^3\)

Hi MWizzard :)

You lost me here

23.1406926328 = i√ i5

you wanted

\(\sqrt[i]{-1}=\ {(-1)}^{1/i}\\ Where\; did\; your\; i^5 \;come\; from?\;\;i^5=i \;\;but\;so\;what?\)

I might be missing something ://

The calculator will show up results like:

x × 10^y or some errors that go above 9.999999999999999 × 10⁹⁹

Also calculators are NOT allowed in the contest of Math Olympiad.

You are welcome.  Why don't you become a member :) 

How can you solve 100x^4-1261x^2+441=0

\(100x^4-1261x^2+441=0\\ let\;\;y=x^2\\ 100y^2-1261y+441=0\\ y=\frac{1261\pm\sqrt{1261^2-4*100*441}}{200}\\ y=\frac{1261\pm 1189}{200}\\ y=\frac{1261+1189}{200}\qquad or \qquad y=\frac{1261-1189}{200} \\ y=\frac{2450}{200}\qquad or \qquad y=\frac{72}{200} \\ y=\frac{1225}{100}\qquad or \qquad y=\frac{36}{100} \\ y=12.25 \qquad or \;\;\quad y=0.36 \\ x^2=12.25 \qquad or \;\;\quad x^2=0.36 \\ x=\pm3.5 \qquad or \;\;\quad x=\pm0.6 \\\)

Solve for x:
100 x^4-1261 x^2+441 = 0

Substitute y = x^2:
100 y^2-1261 y+441 = 0

The left hand side factors into a product with two terms:
(4 y-49) (25 y-9) = 0

Split into two equations:
4 y-49 = 0 or 25 y-9 = 0

Add 49 to both sides:
4 y = 49 or 25 y-9 = 0

Divide both sides by 4:
y = 49/4 or 25 y-9 = 0

Substitute back for y = x^2:
x^2 = 49/4 or 25 y-9 = 0

Take the square root of both sides:
x = 7/2 or x = -7/2 or 25 y-9 = 0

Add 9 to both sides:
x = 7/2 or x = -7/2 or 25 y = 9

Divide both sides by 25:
x = 7/2 or x = -7/2 or y = 9/25

Substitute back for y = x^2:
x = 7/2 or x = -7/2 or x^2 = 9/25

Take the square root of both sides:
Answer: |  x = 7/2    or    x = -7/2    or    x = 3/5    or    x = -3/5

thank you

3 / 5000 = (3 ÷ 50) / (5000 ÷ 50)

              = 0.06 / 100

              = 0.06 %

-1440 feet

negative 1440 feet

We use negative when the number is below 0.

But we use positive when above 0.

intergrate 1/(-0.4v^3)

\(\int \frac{1}{-0.4v^3}\;dv\\ = \frac{1}{-0.4}\int v^{-3}\;dv\\ = \frac{-10}{4}\int v^{-3}\;dv\\ = \frac{-10}{4}\times \frac{v^{-2}}{-2}+c\\ = \frac{5}{4}\times \frac{v^{-2}}{1}+c\\ = \frac{5}{4v^2}+c\\\)